internal combustion engines proje
TRANSCRIPT
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2.61 Internal Combustion Engines
Design Project Corrected Version Number 2
Wednesday, April 14, 2004
Due: Thursday, April 22, 2004
Heavy duty diesel engine with EGR and particulate trap
For this project you need to design a heavy-duty truck diesel engine. The engine is 11
liter, 6 cylinder, with target maximum power of 360 kW and target bsfc of 185 g/kW-hr.You will also incorporate an emissions strategy to meet current emissions standards.
1. Base Engine
Begin by sizing the base engine (no turbocharger), which will be 11 liter, 6 cylinder.
Assume a maximum mean piston speed (Sp) of 10 m/s. Use data on Figure 13.7 on page722 in the text for estimates of mechanical/rubbing plus auxiliary mep. Making any other
reasonable assumptions necessary, determine the following parameters:(a) Bore and stroke
(b) Compression ratio
(c) Connecting rod length(d) Brake mean effective pressure, at maximum torque and maximum power
(e) Maximum torque and maximum power
(f) Maximum engine speed, at maximum power
2. Boost, turbomachinery, and intercooler
Design the required turbomachinery and intercooler for the engine by addressing thefollowing points:
(a) Based on your calculations in part 1, calculate the amount of boost pressurerequired at maximum speed to produce the target power.
(b) Use a turbocharger to produce this boost, and define the main operatingparameters of the required turbomachinery. For both turbine and compressor,
provide values for mass flow rate, pressure ratio, inlet and exhaust temperatures
and pressures. Use typical values for isentropic efficiencies (Kt=0.85, Kc =0.80)and assume the exhaust temperature is 900K.
(c) Include an intercooler to lower the temperature of the air coming out of the
compressor. Provide the inlet and outlet temperatures of the air, as well as the
coolants inlet and outlet temperatures and mass flowrate. Assume a counter-flowheat exchanger with effectiveness of 0.8. (Heat exchanger effectiveness is the
ratio of the actual heat transfer to the heat transfer that would occur if the streamwith the minimum capacity rate were heated (or cooled) from its inlet temperature
to the inlet temperature of the other stream).(d) Draw a schematic of your system(Hint: You will need to include the effect of turbo-charging on pumping work)
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3. Brake efficiencyEvaluate the brake fuel conversion efficiency of the design at half maximum speed and
full load at that speed. Does it meet the target bsfc? If not, what changes in engine designwould bring it closer?
4. Emissions NOXThe engine you are designing must comply with 2004 EPA NOx requirements which are
set at 2.5 g/bhp-hr (assume for simplicity that the NOx requirement must be met at all
operating points); EGR has been chosen as the technology to reduce NOx levels.Address the questions below using available data, and an appropriate safety factor:
(a) Draw a schematic of your system, showing clearly how you will drive EGR fromexhaust to intake. There are a few possibilities for doing this (the reference paper
on EGR systems might be helpful).
(b) Find the required amount of EGR to run at low load (25% max torque and 1600rpm).
(c) Find the required amount of EGR to run at maximum power.
(d) Similar to part 2, recalculate the boost pressure at maximum power. Resize theturbomachinery and intercooler to reach the stated target or best case power
output; use available data to determine the exhaust temperature. Also, include anEGR cooler to lower the re-circulated gas temperature before it enters the engine;
using the same assumptions as part 2c, provide operating temperatures and
flowrates.As a safety measure, it is common standard to reduce the amount of NOx, by anadditional 20% to 40% of the required EPA standard. For this design please use a
safety factor of 30% (i.e., reduce NOx to 1.75 g/bhp-hr, 30% below required
standard). Assume that the equivalence ratio based on the mass of fresh fuel andfresh air must stay below the smoke limit of 0.7; for simplicity once you have
selected the level of EGR, assume NOx levels remain constant, in spite of additionalboosting (this is not the actual case). Also assume that beyond 8 CAD BTDC, for
every additional CAD delay in injection you lose 0.25 percentage points in indicated
fuel conversion efficiency.(Note: Watch the units in the emissions data)
5. Emissions particulate
The engine you are designing must also comply with 2004 EPA particulate requirements.Diesel Particulate Filter (DPF) has been selected as the technology to achieve the target
PM levels of 0.05 g/bhp-hr. Assume that current DPF technology can reach 99%
efficiency (i.e., 99% removal of particulates). Using a PM emissions safety factor of50%, answer the following questions, all for maximum power conditions
(a) What is the approximate level of PM coming out of the engine?
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(b) Size the trap and calculate the average pressure drop. Assume a space velocity inthe range of 10,000 28,000 hr
-1that minimizes the physical size of the
particulate trap (see SAE 2003-01-0047 for reference).
(c) What impact does the particulate trap have on the performance of theturbocharged engine (be quantitative). What changes in boost pressure and
turbomachinery operating conditions are needed to keep best case output?
6. 2007 Emissions requirements:Below is a table showing EPA Diesel engine emissions requirements for 2007.
NOx (g/bhp-hr) PM (g/bhp-hr)
0.20 0.01
(a) Based on engine out NOx levels of part 4, how efficient a NOx catalyst is needed
to meet 2007 emissions levels? Assume that the catalyst is used in conjunctionwith EGR.
(b) Is it possible to achieve these levels of PM with the trap described in the SAE2003-01-0047 paper?
(Note: Keep the same safety factors as in parts 4 and 5)
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E
xhaustTemperaturevs.EGR(ForDifferentStartofInjection@M
aximumPower)
800
850
900
950
1000
1050
5%
10%
15%
20%
25%
30%
35%
%EGR
Temperature(K)
0.5CADBTDC
7.5CADBTDC
3.5CADBTDC
2.5CADATDC
6.5CADBTDC
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25%Maxtorque,1600
RPM
bsNO
xvs.
StartOfMainInjection
Low-LoadCondition
0.0
00
0.5
00
1.0
00
1.5
00
2.0
00
2.5
00
3.0
00
3.5
00
4.0
00
4.5
00
5.0
00
-8
-6
-4
-2
0
2
4
6
8
10
S
tartofMainInjection[CAFromT
DC]
bsNOx[g/kW-hr]
Increase
dEGR
StockEG
R
Reduced
EGR
IncreasingEGR
23.8
%
10.9
%
11.0
%
10.8
%
9.1
%
8.6
%
31.8
%
32.1
%
32.3
%
32.2
%
32.7
%
23.6
%
24.4
%
24.2
%
24.9
%
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bsNOxvs.
Star
tOfMainInjection(Maxim
umP
ower)
0.0
00
1.0
00
2.0
00
3.0
00
4.0
00
5.0
00
6.0
00
7.0
00
8.0
00
9.0
00
-10
-8
-6
-4
-2
0
2
4
6
8
StartofMainInjection[CAFromT
DC]
bsNOx[g/kW-hr]
IncreasedEGR
StockEGR
Reduced
EGR
Increasing
19.9
%
20.6
%
20.4
%
20.2
%
20.0
29.5
30.9
29.6
%
29.5
8.5
%
9.4
%
7.5
%
7.4
7.6
%
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bsPMvs
.StartOfMainInjection
Lo
w-LoadCondition
0.0
000
0.0
200
0.0
400
0.0
600
0.0
800
0.1
000
0.1
200
0.1
400
0.1
600
0.1
800
-8
-6
-4
-2
0
2
4
6
8
10
Start
ofMainInjection[CAFromT
DC]
bsPM[g/kW-hr]
IncreasedEGR
StockE
GR
Reduce
dEGR
Increasing
20.0
22.2
22.9
23.8
23.8
%
31.8
32.1
32.3
32.2
32.7
%
10.9
%
11.0
10.8
9.1
%
8.6
%
25%Maxtorque,1600RP
M
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bsPM
vs
.StartOfMainInjection(MaximumP
ower)
0.0
00
0.0
50
0.1
00
0.1
50
0.2
00
0.2
50
0.3
00
0.3
50
0.4
00
0.4
50
0.5
00
-10
-8
-6
-4
-2
0
2
4
6
8
StartofMainInjection[CAFromT
D
C]
bsPM[g/kW-hr]
Increase
d
EGR
StockEGR
Reduced
EGR
Increasing
8.5
%
9.4
%
7.5
7.4
%
7.6
%
29.5
%
30.9
29.6
29.5
%
20.0
%
20.2
%
20.4
19.9
%
20.6
%
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Displacement (m3) 11
Cylinders 6
Bore (m) 0.1326
Stroke (m) 0.1326
Compression Ratio 18
Connecting Rod Length (m) 0.3315
Bmep @ max torque (kPa) 895
Bmep @ max power (kPa) 835
Maximum torque (N-m) 783
Maximum power (kW) 173
Maximum engine speed at
maximum power(RPM)2261
2.61 Internal Combustion Engines
Design Project Solution
Here is a possible solution for the design problem.
1. Base Engine
Table 1 below summarizes the main parameters of the base engine
Table 1 Base Engine Summary
There are two possible methods to size the engine, and they should be consistent witheach other:
Method 1:
2
/,, stoichoaHVdvifm AFQVNP
IUKKK (1)
Method 2:Assume a bmep based on practical limits and fuel-air cycle charts, and solve for the
power output:
2000
NVdbmepP (2)
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Using the first method, we must determine N,,,,, ,, IUKKK oavifm
Calculate Engine Speed (N)
L2
pSN max find L (3)
Assume B ~ L, so
4
L6
4
B6V
32
d
SS
Lcylinders (4)
mL 1326.06
(0.011)4
6
V4 31
3
1
d
SS
so:
RPM2260or,sec/71.37)1326.0(2
10m/sN revs
Determine I, and Kf,iChose rc=18 (maybe a bit high), and I=0.7 (smoke limit, maximum possible fuel we canget in per mass of air). Using Fuel-air cycle results (Fig. 5-9, Heywood p. 182), then Kf,i=0.575. Applying a correction factor of around 80%, actual Kf,i= 46%. The correctionfactor can be between 80% and 85%; For this case, I chose 80% so that Method 1, and 2,as explained above, are consistent with each other.
Determine IMEP
For phi=0.7, and rc=0.8, we get
5.10imepso5.10imep
1PPi
, (5)
Note that Pi is not atmospheric pressure. At WOT, there is a pressure loss in the intake
system, due to frictional losses that scale with speed. Pi will be less than atmospheric.
Likewise, the exhaust pressure (Pe) is not atmospheric; a higher than atmosphericpressure is needed to pump the gases through the exhaust system. Once the gases leave
the exhaust system and reach ambient conditions, they will expand to atmospheric
pressure. Additionally, depending on the opening timing of the exhaust valves, the gasesmight exit at a higher pressure than what is required to overcome the pumping loss in the
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exhaust system. To get an idea, of the value of Pi, look at Figure 13-13 in the text(Heywood P. 725). For a piston speed of 10 m/s,
(6)
Now allocate this pumping loss between Pe and Pi. At high speeds around 18% of theloss is on the intake side, and the remaining 82% on the exhaust side. This will beconsistent with volumetric efficiency as explained below. So:
Pi= 101 kPa 0.18(40 kPa) = 93.8 kPa
Pe=101 kPa + 0.82(40 kPa) = 133.8 kPa
We can now calculate an imep:
9kPa.9845.10kPa8.93imep
Determine Mechanical Efficiency Km
;1imep
tfmep
imep
tfmepimepm
K (7)
wherepmepfmeptfmpe mep)auxiliaryandfriction(rubbingmepfrictiontotal
From figure 13-7 (Heywood p 722), fmep for a fired engine at 2260 rpm | 140 kPa. So
%7.81985
1801
and;18040140
m
kPakPatfmpe
K
Determine Volumetric Efficiency and oa,U
Using figure 6-8 (Heywood p. 217), assume a volumetric efficiency of 90% for a piston
speed of 10 m/s. Note that this volumetric efficiency measures the efficiency of theentire intake system. Also note that we have chosen the right pressure loss allocation for
the intake system (as calculated in the imep section), consistent with volumetricefficiency. The air density oa ,U , is just calculated from ideal gas law, at ambient
conditions. The value is 1.17 kg/m3
40)10(4.0p)S(4.0Pi)-(Pe 22 xpmep
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Fuel-to-Air Ratio & Heating Value
From table D.4 in the text (Heywood p. 915) we get the stoichiometric Fuel-to-Air ratio
of gasoline as 0.0697, and its heating value of 43.2 MJ/kg
Power calculation
With the estimates for each value, we can now calculate the power
kWP
mkgkgkJemmP
173
2
)0697.0)(7.0(/17.1/32.43)011.0sec)(/7.37)(90.0)(46.0)(817.0( 3
We also use method 2 to check for consistency. Rearranging equation 2.19b (Heywood
p50), we get:
kW
revdmkPaNVdbmepP 167
2000
sec/7.37311805
2000
the methods are close
For low loads, follow the same procedure, with lower pumping loss, due to lower speed
(see figure 13-13, Heywood), and lower rubbing and auxiliary friction (see figure 13-7Heywood); additionally, the allocation of pressure losses is different, and must be
consistent with volumetric efficiency.
2. Boost, Turbo-machinery and Intercooler
Boost pressure:
To find the boost pressure required, we use equation 1, and replace the volumetricefficiency for the entire inlet system with the volumetric efficiency for the valves only
( vK ~ 94%). We also replace the ambient air density with the air density right before the
valves, ia,U . This density can be determined from the ideal gas law, knowing the pressure
(which is approximately cylinder pressure divided by volumetric efficiency), and thetemperature (about the same as the cylinder temperature). Thus, we can vary the cylinder
pressure until we get the required power level, as defined by equation 1.
2
/,, stoichiaHVdvifm AFQVNP
IUKKK
Note that as we vary the cylinder pressure, and consequently the density, the mechanical
efficiency (as defined by equation 7 above) will also change because the pumping loss
will change.
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Pmep= Pexhaust - Pintake
Thus the solution to this problem is iterative, and can easily be done with a spreadsheet.
After varying the cylinder pressure, determining the corresponding air density at thevalves through the ideal gas law, and calculating the mechanical efficiency, we get the
following target power:
kWP
mkgkgkJemmP
360
2
)0697.0)(7.0(/065.2/32.43)011.0sec)(/7.37)(944.0)(46.0)(918.0( 3
For this case the pressure that gives a density of 2.065 is 176 kPa, as dictated by the ideal
gas law:
)944.0(314/97.28/314.8
/065.2* 3_)()(, Kkmolekg
kmoleKkJmkgRTP valvesvesbeforevalvesbeforevalviacylinder
KU
which gives Pcylinder=176 kPa. The pressure that must come out of the compressor is
approximately:
kPakPaP
Pv
cylinder
comp 186944.0
176
K
Thus the desired boost is 85 kPa. That is we have to compress 85 kPa aboveatmospheric. Note that to relate pressure before the valves, and after the valves, as a first
approximation I have used the volumetric efficiency.
Turbo-machinery
Knowing the desired boost, the turbo-machinery can now be sized to generate the
required pressure. This is done using the insentropic relationships for the compressor andturbine. First we must size the compressor by finding the work required to compress the
gas to the desired pressure. Second, we must size the turbine to produce the work that
drives the compressor.
To determine the amount of work that is required to compress the gas we do an energybalance assuming an adiabatic compressor:
)( 12 TTCmW apc (8)
where,
T2a= Actual compressor exit temperatureT1= Compressor inlet temperature (300K)
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We calculate T2a using the compressor efficiency, and isentropic relationships:
112
2 TTT
Tc
sa
K(9)
and:
J
J1
2
112
P
PTTs (10)
T1, P1, P2, J, and Kc are all known, so T2a, and consequently the compressor work canbe calculated.
Knowing the compressor work, we now size the turbine using the following equations:
m
ct
WW
K (11)
where mK is the mechanical efficiency for the turbine and compressor system. 95% isreasonable estimate for this number
Cp
WTT ta 45 (12)
where:T5a=Actual turbine exhaust temperature
T4 = Turbine inlet temperature (engine exhaust, given at 900K)
To find the required turbine pressure ratio:
J
J
1
4
5
5
4
T
TP
P
P sr (13)
where:
445
5 TTT
Tt
as
K(14)
Thus, enough equations for enough unknowns. Values for the temperatures, pressures,
and compressor work, are show in table 2.
Intercooler
Adding an intercooler to lower the intake temperature, will increase the density of the
gas, and consequently decrease the required boost, as reflected in table 2. For a given
pressure rise we get a higher change in density (due to lower gas temperatures going intothe engine). To size the intercooler you can select a coolant, and based on adequate
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compressor turbine
Heat Exchanger Engine
1
2 3
4
5
67
estimates for inlet and outlet coolant temperatures, you can determine the required massflow-rate that is needed to achieve a certain temperature change in the air. You must use
the definition for heat exchanger to determine the allowed change in air temperature:
For the coolant I used water (Cp=4.2 kJ/kg K), and assumed that it goes in at 300 K, andI want it to leave at 380 K.
To find the exit temperature of the air, and to determine the required flow rate of water, Iuse the definition for heat exchanger effectiveness in conjunction with an energy balance:
For effectiveness we have:
)(
)(
max_min
airor,
outin
coolantoutinairorcoolant
TTCpm
TTCpm
H
and for the energy balance we have:
aircoolant TCpmTCpm ''
For this case, I chose the air and water to have about the same capacitance (mCp). Using
the effectiveness equation I can solve for Tout air. Note that the capacitances will cancelout in the equation, and the maximum change in temperature occurs when Tout air=
Twater in, thus:
)()(
)(____
__
irr
inwaterinairinairoutair
inwaterairin
aoutin TTTTTT
TT
HH
Assuming, water temperature increases from 300 to 357, then Tair in is 314 K. Valuesfor the heat exchanger temperatures and flow-rates are also shown in table 2.
Figure 1
Schematic of Turbocharged Engine with Intercooler
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State Temperature (K) Pressure (kPa) Temperature (K) Pressure (kPa)
1 300 101 300 1012 401 233 372 186
3 401 233 314 186
4 900 223 900 188
5 801 127 833 127
6 N/A N/A 101 300
7 N/A N/A 101 380
Work Compressor 40 kW Work Compressor 29 kW
Intercooler
m_dot water 0..097 kg/sec
m_dot air 0.404 kg/sec
No intercooling With intercooling
Table 2
Turbocharged Engine with Intercooler: Operating Parameters
3. Brake Efficiency
At half maximum speed, and full load at that speed, I kept the same boost, but loweredthe fmep, per figure 13-7 in the text. The BSFC came out to be 188 g/kw-hr. This
number is actually quite good for industry standards. Other people perhaps got lower
(around 175), however, as I previously explained, I was more conservative in myefficiency estimate from fuel air cycle tables, to be consistent with different ways of
calculating power. My calculation is shown below:
hrkWgkgghrkg
kWPower
hrgmbsfc
f /188193
)/1000)(sec/3600sec(/010078.0
)(
)/(
Note that if we directly use break engine efficiency, we should get the same answer:
hrkWgQQ
bsfcHVifmHVbf
/188)2.43)(460.0(96.0
1)(
1)(
1
,, KKK
Ways to decrease bsfc include raising compression ratio, and reducing frictional losses.
4. Emissions NOx
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compressor turbine
Inter-
coolerEngine
1
2 3
5
6
Inter-
coolerVenturi/
Mixer
4
7
Engine ModeNOx Standard
g/bhp-hr
NOx safety
target g/bkW-
hr
NOx Safegy
target g/bkW-
hr
EGRTiming (CA
from TDC)
Hit in Fuel
economy
(percentage
points)
25% maxtorque, 1600
rpm
2.50 1.75 2.35 24% 1 2%
Maximum
power2.50 1.75 2.35 24% 0.5 2%
Figure 2
Schematic of Turbocharged Engine with Intercooler
The schematic shows how EGR will be driven from the engine. There are a few ways of
doing this; one way is to use a Venturi system, as shown above. Another way is tooptimize the system so that the pressures at the air and EGR intersection are about thesame. It is necessary for these pressures to be equal, otherwise there will be backflow in
the direction of lower pressure. Overall, however, the addition of EGR will impact the
fuel economy of the engine. This is the price that we must pay to have lower emissions.
The first step of this problem is to define the amount of EGR that is needed to meet EPAemissions levels. The emissions requirements along with their safety levels are shown in
table 3 below.
Table 3
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Must operate
below dashed
line
Using the figures provided, there are various possibilities for selecting EGR, depending
on the hit on fuel economy. Figure 3 below is an example that shows that there is a rangeof timings and EGR levels that will give the proper amount of NOx
Figure 3
Acceptable operating area for low load
Once EGR has been calculated, the loss in engine efficiency can be assessed, as well as
the required boost. Again this is an iterative process. There are many variables affectingengine power, and they are all related as well, thus at least a spreadsheet must be setup.
For example, boost affects engine power, but it also affects mechanical efficiency, which
in turn affects engine power, thus all these variables must be connected when solving thesystem.
One important implication of adding EGR, is that the pressure in the cylinder chamber
must increase if we are to maintain constant mass of fresh fuel and air; this is what we
should desire if we are to maintain the same power output from the engine as the casewithout EGR.
The total pressure is equal to the sum of the partial pressures of air and the EGR:
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State Temperature (K) Pressure (kPa)
1 300 101
2 434 295
3 327 295
4 349 279
5 990 241
6 864 121
7 438 241
With intercooling
IntercoolerMass flowrate
(kg/sec)Tin (K) Tout (K)
EGR 0.171 300 380
Compressor 0.131 300 380
EGRairT PPP
Assuming the molecular weights of both Air and EGR are about the same, then the mole
fraction is approximately equal to the mass fraction of each mixture, and PT can beexpressed as:
EGR
PP airT
1
Additionally, since there is a pressure loss of around 16 to 20 kPa associated with the
venturi, a higher boost is still needed. To reach the target power output, a total boost of
194 kPa is required, for total pressure of 295 kPa. This is a high boost, higher thanindustry standard for this size engines. Perhaps a more practical boost is 150 kPa (PT=251 kPa), or less. However this limits the maximum power to 310 kW. If yourecognized the practical limitations, this is a perfectly acceptable answer.
Table 4
Table 5
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5. Emissions Particulates
(a) The particulate emissions corresponding to the chosen EGR level, can be obtained
from the data provided (PM levels vs injection timing for various EGR fractions). At
24% EGR, the PM coming out of the engine is approximately 0.2 g/bkW-hr
(b) To size the trap, use the space velocity that will minimize the volume of the trap, inthis case 28,000 hr-1
. This is evident from the relationship for space velocity:
velocitySpaceV
V_
where V is the volume flow-rate of the gases going through the trap, and V is the volume
of the trap. For a smaller trap volume we get a higher space velocity.
Solving for the volume of the trap we get
velocitySpace
VV
_
Using the ideal gas law to solve forV
sec/10.1)(
3mP
RTmmmV
exhaust
exhaustegrfuelair
Solving for Volume
LmmV 141141.0sec3600/000,28
sec/10.1 33
Values used are shown in table 6 below
Table 6
As shown in figure 4 of SAE 2003-01-0047, The maximum pressure loss through the
trap, at a space velocity of 28,000/hr, is 6 kPa. This is a very small percentage of thetotal exhaust pressure (~2.5%), and the effect on turbo-machinery is small.
Mdot_air&fuel
(kg/sec)
Mdot_egr
(kg/sec)Texhaust (K) Pexhaust (kPa)
Space Velocity
(1/sec)
Volume
flowrate
(m3/sec)
Trap Volume
(L)
0.43 0.10 864.74 121.00 7.78 1.10 140.87
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6. 2007 Emissions requirements
Using the same safety factor as in part 6, the table below shows the new emissions that
must be met:
Table 7
Where catalytic converter efficiency is defined as:
intpollu
outtpollu
catm
m
,tan
,tan1
K
As shown in table 7, a catalytic converter with 92% efficiency will be needed to meet
2007 NOx emissions requirements.
The required particulate trap efficiency is fairly high (97%) but the trap presented in the
Ford paper seems to have efficiencies of around 99%, so it should work fine for 2007emissions requirements.
NOx
Standard
g/bhp-hr
NOx safety
target
g/bhp-hr
NOx
Safegy
target
g/bkW-hr
Current
Engine out
g/bkW-hr
Required
efficiency
(catalytic)
PM
Standard
g/bhp-hr
PM safety
target
g/bhp-hr
PM Safegy
target
g/bkW-hr
Current
Engine out
g/bkW-hr
Required
Efficiency
(trap)
0.2000 0.1400 0.1879 2.3490 0.9200 0.0100 0.0050 0.0067 0.2000 0.966